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Let $\left(Z, *\right)$ be an algebraic structure where $Z$ is the set of integers and the operation $*$ is defined by $n*m = \max(n,m)$. Which of the following statements is true for $\left(Z, *\right)$?

1. $\left(Z, *\right)$ is a monoid
2. $\left(Z, *\right)$ is an Abelian group
3. $\left(Z, *\right)$ is a group
4. None of the above

### 4 Comments

what about a*a = max(a, a)?How it can be an algebric structure if max value of the same elements does not exists. Please correct me if i am wrong
If the set was for Natural numbers we could have had 1 as identity element as it is less than any other natural number and we would have got a monoid.

what about max(1,1)=??
Hello sushmita

What is restraining you from considering $max(n,n)=n$ ?
If set would have been of +ve integers then we've 1 as identity right...?

## 6 Answers

Best answer

Lets follow our checklist one by one to see what property this algebraic structure follows.

Closure -yes ($m*n=\max(m,n)$) Output is either $m$ or $n$ whichever is maximum and since $m,n$ belongs to $Z$, the result of the binary operation also belongs to $Z$. So closure property is satisfied.

Associative-Yes the output is max among the elements and it is associative

Now for identity, we don't have a single unique element for all the elements which is less than all the elements. ie, $m*e=m \implies \max(m,e) = m$. We can't find a single unique $e$ which is less than all possible integer $m$, such that comparison between the two would always give $m$ itself.

If the set was for Natural numbers we could have had 1 as identity element as it is less than any other natural number and we would have got a monoid.

• Semi-group - Closed and associative
• Monoid - Closed, associative and has an identity
• Group - Monoid with inverse
• Abelian group- Group with commutative property.

Hence, this is just a semigroup- D option.

Ans D)

by

### 4 Comments

Z is set of integers means { -infinity........-2, -1, 0, 1, 2........infinity}, not like set of natural numbers {1,2,3,4,......infinity}

How can u find the smallest element in Z ???

that's why Z is not a monoid

@ Relax. No need to be angry :p

-INF is not a number so it does not belong to our set.

Given (Z,*) is an algebraic structure. Hence it is closed.

Now, to become SEMI-GROUP: It should be associative

(x*y)*z = max(max(x,y),z) = max(x,y,z)

x*(y*z) = max(x,max(y,z)) = max(x,y,z)

eg. (1*2)*3 = max((max(1,2),3) = max(2,3) = 3

1*(2*3) = max(1,max(2,3)) = max(1,3) = 3

Hence (x*y)*z = x*(y*z), so ASSOCIATIVE i.e SEMI-GROUP

Now, to become MONOID: It should have an identity element

Let there be such an element and let the identity element be e.

So, as per definition: (a*e) = max(a,e) = a

Nice. But the question is: What if we take e=(a+1) ∀ a ∈ Z

Then, (a*(a+1)) = max(a,(a+1)) = (a+1) and NOT a

So, there can be no such identity(e) element possible.

Hence, its not a MONOID.

So correct choice is: (D) None

Closure Property => This is satisifed .Maximum of two integer is one of the integers. So we satisfy closure property.

Associativity => This property also satisfied.

2,3,4 =Max(2(max(3,4)) = 4 = Max(Max(2,3),4) .

Identity => We do not have identity.

Suppose some i is identity.

Then Max(i,anything) should be that anything !

But if I  take i-1. (As we have Z , If i is integer so is i-1)

Max(i,i-1) = i. So i is not identity.

So This is not monoid. & SO not group or abelian group . Ans is D
Hi , in this question does identity element always exist ? because , if I take a set A = {-2,-1,0,1,2}

( 0 can be included as it is integer set ).

so , -2 * e = -2 ( where e is the identity element ) , but if here we can not have any such unique identity element .

So , this is not going to be Monoid .

Please correct me , if I am wrong.
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